Question

A convex lens forms a real image 4 times magnified at a distance of 60 cm from the lens. Calculate the focal length and the power of the lens. class 10 phy​

Answer 1

Answer:

Explanation:

Magnification (m) = v/u

-4 = 60/u

So, u = 60/-4 = -15cm

1/f = 1/v - 1/u = 1/60+1/15 = 5/60

Thus, f = 60/5 = 12 cm = 0.12 m

P = 1/f = 1/0.12 = 8.33 D

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Answer 2

Given :-

• Magnification produced by the lens, m = -4 (for real image).
• Image distance, v = 60 cm.

To Find :-

• Focal length and power of the lens.

Solution :-

We know, magnification produced by a lens is related to the object distance (u) and the image distance (v) as follows:-

$\implies\rm{m=\dfrac{v}{u}{\implies}u=\dfrac{v}{m}}$

$\implies\rm{u=\dfrac{60}{-4}=-15\:cm}$

Now, we have object distance & image distance. Then, by using lens formula,

$\implies\rm{\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}}$

We get,

$\implies\rm{\dfrac{1}{f}=\dfrac{1}{60}-\dfrac{1}{-15}=\dfrac{1+4}{60}}$

$\implies\rm{\dfrac{1}{f}=\dfrac{5}{60}=\dfrac{1}{12}}$

$\implies\rm{f=}\:\bold{12\:cm.}$

$\rule{350}{3}$

Now, power of the lens, $\rm{P=\dfrac{1}{f\:(in\:m)}}$

$\implies\rm{P=\dfrac{1}{\dfrac{12}{100}}}$

$\implies\rm{P=\dfrac{100}{12}D=}\:\bold{8.33\:D.}$